📚 Year 10 AQA Chemistry: Quick Reference Handbook of Formulas and Theorems | Year 10 AQA 化学：公式定理速查手册
Mastering key formulas and principles is essential for success in AQA GCSE Chemistry. This handbook covers all the calculations and core concepts encountered in Year 10, from the mole and concentration to energy changes and electrolysis. Use it as a revision companion to reinforce your understanding and solve problems accurately.
掌握关键公式和原理是 AQA GCSE 化学取得成功的必要条件。本手册涵盖了 Year 10 学习的所有计算和核心概念，从摩尔和浓度到能量变化和电解。将其用作复习伴侣，以巩固理解并准确解题。
1. Atomic Structure & Relative Atomic Mass | 原子结构与相对原子质量
Atoms consist of a central nucleus (protons + neutrons) surrounded by electrons arranged in energy levels (shells). The number of protons defines the element (atomic number). The mass number is the total number of protons and neutrons, while isotopes are atoms of the same element with different numbers of neutrons.
原子由中心原子核（质子+中子）和核外按能级（电子层）排列的电子组成。质子数决定了元素（原子序数）。质量数是质子数与中子数之和，而同位素是同一元素中中子数不同的原子。
Relative atomic mass, Aᵣ, is the weighted average mass of the isotopes of an element relative to 1/12th the mass of a carbon-12 atom. The formula for Aᵣ from isotopic data:
相对原子质量 Aᵣ 是某元素同位素原子质量的加权平均值，相对于碳-12原子质量的1/12。根据同位素数据计算 Aᵣ 的公式为：
Aᵣ = (Σ abundance % × mass number) ÷ 100
For chlorine with 75% of ³⁵Cl and 25% 25 × 37) ÷ 100 = 35.5. Always check that your answer lies between the two isotope masses.
以氯为例，75% 为 ³⁵Cl、25% 为 ³⁷Cl：Aᵣ = (75 × 35 + 25 × 37) ÷ 100 = 35.5。务必检查答案是否落在两个同位素质量数之间。
2. Electronic Configuration & the Periodic Table | 电子排布与元素周期表
Electrons occupy shells that fill in the order 2, 8, 8 (and then 2 for the first 20 elements). The group number equals the number of electrons in the outer shell, and the period number equals the number of occupied shells. Elements in the same group therefore have similar chemical properties.
电子依次填入电子层，顺序为 2、8、8（前 20 号元素再补 2）。主族序数等于最外层电子数，周期数等于被占用的电子层数。因此同族元素具有相似的化学性质。
- Group 1 (alkali metals): 1 outer electron, form 1⁺ ions, reactivity increases down the group.
第 1 族（碱金属）：最外层 1 个电子，形成 1⁺ 离子，自上而下活泼性增强。 - Group 7 (halogens): 7 outer electrons, form 1⁻ ions, reactivity decreases down the group.
第 7 族（卤素）：最外层 7 个电子，形成 1⁻ 离子，自上而下活泼性减弱。 - Group 0 (noble gases): full outer shell, chemically inert, boiling points increase down the group.
第 0 族（稀有气体）：最外层已满，化学性质稳定，沸点自上而下升高。
3. Bonding: Ionic, Covalent and Metallic | 化学键：离子键、共价键与金属键
Ionic bonding involves the transfer of electrons from a metal to a non-metal, producing oppositely charged ions held in a giant lattice. Covalent bonding involves the sharing of electron pairs between non-metal atoms. Metallic bonding consists of positive ions in a sea of delocalised electrons.
离子键是金属原子将电子转移给非金属原子，形成带相反电荷的离子并构成巨型晶格。共价键是非金属原子之间共用电子对。金属键由阳离子与自由电子海构成。
The formula of an ionic compound is found by balancing the charges so that the total is zero:
离子化合物的化学式通过使电荷总和为零来配平得出：
| Ions | Charge balance | Formula |
| Na⁺ and Cl⁻ | (+1) + (−1) = 0 | NaCl |
| Mg²⁺ and Cl⁻ | (+2) + 2(−1) = 0 | MgCl₂ |
| Al³⁺ and O²⁻ | 2(+3) + 3(−2) = 0 | Al₂O₃ |
Substances with giant structures (ionic, covalent or metallic) have high melting points; simple molecular substances have weak intermolecular forces and low melting points.
具有巨型结构的物质（离子、共价或金属）熔点高；简单分子物质分子间作用力弱，熔点低。
4. Relative Formula Mass & Percentage Composition | 相对式量与质量分数
The relative formula mass, Mᵣ, is the sum of the relative atomic masses of every atom shown in the formula. It has no units and is used for both ionic and covalent compounds.
相对式量 Mᵣ 是化学式中所有原子相对原子质量的总和。它没有单位，既可用于离子化合物也可用于共价化合物。
Mᵣ = Σ (Aᵣ of each atom × number of atoms)
% mass of element = (Aᵣ × number of atoms ÷ Mᵣ) × 100
r, H₂O: Mᵣ = (2 × 1) + 16 = 18. The percentage by mass of oxygen is (16 ÷ 18) × 100 = 88.9%. For CO₂: Mᵣ = 12 + (2 × 16) = 44, and oxygen is (32 ÷ 44) × 100 = 72.7%.
以水 H₂O 为例：Mᵣ = (2 × 1) + 16 = 18，氧元素的质量分数为 (16 ÷ 18) × 100 = 88.9%。对 CO₂：Mᵣ = 12 + (2 × 16) = 44，氧为 (32 ÷ 44) × 100 = 72.7%。
5. The Mole and Avogadro's Constant | 摩尔与阿伏加德罗常数
A mole is the amount of substance that contains 6.02 × 10²³ particles (Avogadro' the mass you can weigh in the laboratory and the number of particles that actually react.
摩尔是含有 6.02 × 10²³ 个粒子（阿伏加德罗常数）的物质的量。摩尔是实验室中可称量的质量与实际参加反应的粒子数之间的桥梁。
moles (mol) = mass (g) ÷ Mᵣ
number of particles = moles × 6.02 × 10²³
× 10²² molecules. Rearranging the triangle gives mass = moles × Mᵣ and Mᵣ = mass ÷ moles.
例：4.4 g 二氧化碳含 4.4 ÷ 44 = 0.1 mol，即 0.1 × 6.02 × 10²³ = 6.02 × 10²² 个分子。将公式三角形变形可得：质量 = 摩尔数 × Mᵣ，Mᵣ = 质量 ÷ 摩尔数。
6. Conservation of Mass & Reacting Mass Calculations | 质量守恒与反应质量计算
Mass is conserved in a chemical reaction because atoms are rearranged, not created or destroyed. A balanced equation gives the mole ratio in which substances react, which is the key to every reacting mass calculation.
化学反应中质量守恒，因为原子只是重新排列，既不会产生也不会消失。配平的化学方程式给出各物质反应的摩尔比，这是所有反应质量计算的关键。
- Step 1: write and balance the equation.
第一步：写出并配平化学方程式。 - Step 2: convert the known mass into moles using n = m ÷ Mᵣ.
第二步：用 n = m ÷ Mᵣ 将已知质量换算为摩尔数。 - Step 3: use the mole ratio from the equation to find the moles of the unknown.
第三步：利用方程式中的摩尔比求出未知物质的摩尔数。 - Step 4: convert moles back into mass using m = n × Mᵣ.
第四步：用 m = n × Mᵣ 把摩尔数换算回质量。
, n = 4.8 ÷ 24 = 0.2 mol, so n(MgO) = 0.2 mol and m = 0.2 × 40 = 8.0 g of magnesium oxide.
例：2Mg + O₂ → 2MgO。4.8 g 镁：n = 4.8 ÷ 24 = 0.2 mol，故 n(MgO) = 0.2 mol，m = 0.2 × 40 = 8.0 g 氧化镁。
7. Concentration of Solutions | 溶液浓度
Concentration tells you how much solute is dissolved in a given volume of solution. It can be expressed in g dm⁻³ or, more usefully for calculations, in mol dm⁻³.
浓度表示一定体积溶液中所含溶质的量，可用 g dm⁻³ 表示，而在计算中更常用 mol dm⁻³。
concentration (mol dm⁻³) = moles ÷ volume (dm³)
concentration (g dm⁻³) = mass (g) ÷ volume (dm³)
e relationship gives n = c × V, and the balanced equation converts moles of one substance into moles of the other.
记住 1 dm³ = 1000 cm³，因此 250 cm³ = 0.25 dm³。例：250 cm³ 溶液中有 0.05 mol 溶质，浓度为 0.05 ÷ 0.25 = 0.2 mol dm⁻³。在滴定中，同一关系式给出 n = c × V，再通过配平的方程式将一种物质的摩尔数换算为另一种。
8. Gas Volumes at Room Temperature and Pressure | 室温常压下的气体体积
At room temperature and pressure (RTP, 20 °C and 1 atm), one mole of any gas occupies 24 dm³. This molar gas volume lets you link the volume of a gas directly to the number of moles.
在室温常压（RTP，20 °C、1 atm）下，任何气体 1 摩尔都占据 24 dm³。这一摩尔气体体积使气体体积与摩尔数直接关联。
moles of gas = volume (dm³) ÷ 24
× 10²³ molecules. Always convert cm³ to dm³ by dividing by 1000 before using the formula.
例：RTP 下 12 dm³ 氢气为 12 ÷ 24 = 0.5 mol，含 0.5 × 6.02 × 10²³ = 3.01 × 10²³ 个分子。使用公式前务必先把 cm³ 除以 1000 换算为 dm³。
9. Percentage Yield, Atom Economy and Purity | 产率、原子经济性与纯度
Percentage yield compares the mass actually obtained with the maximum theoretical mass. Atom economy compares the mass of the desired product with the total mass of all reactants, and is a measure of how efficiently atoms are used.
产率将实际得到的质量与理论最大质量相比较；原子经济性将目标产物的质量与所有反应物的总质量相比较，衡量原子利用的效率。
% yield = (actual mass ÷ theoretical mass) × 100
% atom economy = (Mᵣ of desired product ÷ Σ Mᵣ of all reactants) × 100
% purity = (mass of pure substance ÷ total mass of sample) × 100
if the theoretical mass is 5.0 g and you obtain 4.5 g, the yield is (4.5 ÷ 5.0) × 100 = 90%. Yield is always below 100% because of incomplete reactions, side reactions and loss during separation.
例：理论质量 5.0 g，实际得到 4.5 g，则产率为 (4.5 ÷ 5.0) × 100 = 90%。产率总低于 100%，原因是反应不完全、发生副反应以及在分离过程中损失。
10. Energy Changes: Bond Energies and Q = mcΔT | 能量变化：键能与 Q = mcΔT
Exothermic reactions release energy to the surroundings and show a negative ΔH; endothermic reactions absorb energy and show a positive ΔH. During a reaction, bonds in the reactants are broken (energy in) and new bonds in the products are formed (energy out).
放热反应向环境释放能量，ΔH 为负；吸热反应吸收能量，ΔH 为正。反应中反应物的化学键断裂（吸能），生成物的新键形成（放能）。
ΔH = energy needed to break bonds − energy released forming bonds
Q = m × c × ΔT
n joules. Example for H₂ + Cl₂ → 2HCl: bonds broken = 436 + 243 = 679 kJ, bonds formed = 2 × 431 = 862 kJ, so ΔH = 679 − 862 = −183 kJ mol⁻¹, an exothermic reaction.
在 Q = m × c × ΔT 中，m 为水的质量（g），c 为比热容（水为 4.18 J g⁻¹ °C⁻¹），ΔT 为温度变化（°C），Q 的单位为焦耳。以 H₂ + Cl₂ → 2HCl 为例：断键吸能 = 436 + 243 = 679 kJ，成键放能 = 2 × 431 = 862 kJ，故 ΔH = 679 − 862 = −183 kJ mol⁻¹，为放热反应。
11. Electrolysis and Charge Calculations | 电解与电荷计算
Electrolysis uses a direct current to decompose an ionic compound, either molten or in aqueous solution. Positive ions (cations) move to the cathode and are reduced; negative ions (anions) move to the anode and are oxidised.
电解利用直流电分解熔融或水溶液状态的离子化合物。阳离子移向阴极被还原，阴离子移向阳极被氧化。
Q = I × t
moles of electrons = Q ÷ 96 500
which is 7200 ÷ 96 500 = 0.075 mol of electrons. Since Al³⁺ + 3e⁻ → Al, this deposits 0.075 ÷ 3 = 0.025 mol of aluminium.
Q 为电荷量（库仑），I 为电流（安培），t 为时间（秒）。例：2 A 电流通电 1 小时（3600 s），Q = 2 × 3600 = 7200 C，即 7200 ÷ 96 500 = 0.075 mol 电子。由 Al³⁺ + 3e⁻ → Al 可知，可析出 0.075 ÷ 3 = 0.025 mol 铝。
12. Rates of Reaction and Equilibrium | 反应速率与化学平衡
The rate of a reaction is the change in amount of reactant or product per unit time. It increases with higher temperature, higher concentration or pressure, greater surface area, and the presence of a catalyst, because these factors increase the frequency of successful collisions.
反应速率是单位时间内反应物或生成物的变化量。升高温度、增大浓度或压强、增大接触面积以及使用催化剂都能提高反应速率，因为这些因素增加了有效碰撞的频率。
mean rate = amount of reactant used or product formed ÷ time
On a graph of product against time, the rate at any point is the gradient of the curve; the steepest gradient is the initial rate, and the gradient becomes zero when the reaction stops. In a reversible reaction at equilibrium, the forward and reverse rates are equal and the concentrations remain constant; changing temperature, pressure or concentration shifts the position of equilibrium.
在产物–时间图像中，任一点的速率即该点曲线的斜率；斜率最大处为初始速率，反应结束时斜率变为零。在可逆反应达到平衡时，正逆反应速率相等，各物质浓度保持不变；改变温度、压强或浓度会使平衡发生移动。
Keep this handbook beside your exercise book, cover the formulas and test yourself weekly, and always show the equation, the substitution and the unit in every calculation answer. Working through past-paper questions under timed conditions is the fastest way to turn these formulas into marks.
把本手册放在练习册旁，遮住公式每周自测一次，并在每道计算题中写出方程式、代入过程和单位。在限时条件下练习历年真题，是把这些公式转化为分数的最快途径。
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